\appendix{命题\ref{Prn:P2P:SISO:Lower:TG=ABG}的证明}\label{Apx:P2P:SISO:Lower:TG=ABG}
    当$ -A\leq X\leq A $时，有
\begin{align}
f^{\mathrm{TG}}_X\left(x\right)=e^{-1-\left(\frac{u^2}{2v^2}+\ln{\eta v\sqrt{2\pi}}-1\right)-\left(-\frac{u}{v^2}\right)x-\frac{1}{2v^2}x^2}.
\end{align}

为了便捷，本文定义如下三个变量
\begin{subequations}\label{Eqn:Apx:P2P:SISO:Lower:TG=ABG:widetiled_val}
    \begin{align}
    &\widetilde{\alpha} \triangleq \frac{u^2}{2v^2}+\ln{\eta v\sqrt{2\pi}}-1,\\
    &\widetilde{\beta} \triangleq -\frac{u}{v^2},\\
    &\widetilde{\gamma} \triangleq \frac{1}{2v^2}.
    \end{align}
\end{subequations}

将\eqref{Eqn:Apx:P2P:SISO:Lower:TG=ABG:widetiled_val}代入\eqref{Eqn:P2P:SISO:Lower:MaxEntropy:Solution:a}等号左侧，得
\begin{subequations}
    \begin{align}
    &\frac{\sqrt{\pi}e^{\frac{\widetilde{\beta}^2}{4\widetilde{\gamma}}}\left(\erf{\frac{\widetilde{\beta}+2\widetilde{\gamma} A}{2\sqrt{\widetilde{\gamma}}}}-\erf{\frac{\widetilde{\beta}-2\widetilde{\gamma} A}{2\sqrt{\widetilde{\gamma}}}}\right)}{2\sqrt{\widetilde{\gamma}}e^{1+\widetilde{\alpha}}}\\
    &=\frac{\sqrt{\pi}e^{\frac{\left(\tgB\right)^2}{\frac{2}{v^2}}}\left(\erf{\frac{\tgB+\frac{ A}{v^2}}{2\sqrt{\tgG}}}-\erf{\frac{\tgB-\frac{ A}{v^2}}{2\sqrt{\tgG}}}\right)}{2\sqrt{\tgG}\expOnetgA}\\
    &=\frac{\erf{\frac{A-u}{v\sqrt{2}}}-\erf{\frac{-A-u}{v\sqrt{2}}}}{2\eta}\\
    &=\frac{\left(\frac{1}{2}+\frac{1}{2}\erf{\frac{A-u}{v\sqrt{2}}}\right)-\left(\frac{1}{2}+\frac{1}{2}\erf{\frac{-A-u}{v\sqrt{2}}}\right)}{\eta}\\
    &=\frac{\Phi\left(\frac{A-u}{v}\right)-\Phi\left(\frac{-A-u}{v}\right)}{\eta}\\
    &=1.
    \end{align}
\end{subequations}

将\eqref{Eqn:Apx:P2P:SISO:Lower:TG=ABG:widetiled_val}代入\eqref{Eqn:P2P:SISO:Lower:MaxEntropy:Solution:b}等号左侧，得
\begin{subequations}
    \begin{align}
    &\frac{1}{2\widetilde{\gamma} e^{1+\widetilde{\alpha}}}\left(e^{A\left(\widetilde{\beta}-\widetilde{\gamma} A\right)}-e^{-A\left(\widetilde{\beta}+\widetilde{\gamma} A\right)}\right)-\frac{\widetilde{\beta}}{2\widetilde{\gamma}}\\
    &=\frac{v^2}{\expOnetgA}\left(e^{A\left(\tgB-\frac{A}{2v^2} \right)}-e^{-A\left(\tgB+\frac{A}{2v^2}\right)}\right)-\frac{\tgB}{\frac{2}{2v^2}}\\
    &=\frac{v}{\eta}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{A^2+2Au+u^2}{2v^2}}-\frac{1}{\sqrt{2\pi}}e^{-\frac{A^2-2Au+u^2}{2v^2}}\right)+u\\
    &=\frac{v}{\eta}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(-A-u\right)^2}{2v^2}}-\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(A-u\right)^2}{2v^2}}\right)+u\\
    &=\frac{v}{\eta}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{-A-u}{v}\right)^2}{2}}-\frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{A-u}{v}\right)^2}{2}}\right)+u\\
    &=\frac{v}{\eta}\left(\pdfl-\pdfu\right)+u\\
    &=0.
    \end{align}
\end{subequations}

将\eqref{Eqn:Apx:P2P:SISO:Lower:TG=ABG:widetiled_val}代入\eqref{Eqn:P2P:SISO:Lower:MaxEntropy:Solution:c}等号左侧，得
\begin{subequations}
    \begin{align}
    &\frac{1}{4\widetilde{\gamma}^2 e^{1+\widetilde{\alpha}}}\left[\left(\widetilde{\beta}-2\widetilde{\gamma} A\right)e^{-A\left(\widetilde{\beta}+\widetilde{\gamma} A\right)}-\left(\widetilde{\beta}+2\widetilde{\gamma} A\right)e^{A\left(\widetilde{\beta}-\widetilde{\gamma} A\right)}\right]+\frac{\widetilde{\beta}^2+2\widetilde{\gamma}}{4\widetilde{\gamma}^2}\\
    &=\frac{v^4}{\expOnetgA}\left[\left(\tgB-\frac{A}{v^2}\right)e^{-A\left(\tgB+\frac{A}{2v^2}\right)}-\left(\tgB+\frac{A}{v^2}\right)e^{A\left(\tgB-\frac{A}{2v^2}\right)}\right]+\frac{\left(\tgB\right)^2+\frac{1}{v^2}}{\frac{1}{v^4}}\\
    &=\frac{v^2}{\eta}\left[\svall \frac{1}{\sqrt{2\pi}}e^{-\frac{A^2-2Au+u^2}{2v^2}}-\svalu \frac{1}{\sqrt{2\pi}}e^{-\frac{A^2+2Au+u^2}{2v^2}} \right]+u^2+v^2\\
    &=\frac{v^2}{\eta}\left[\svall \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{A-u}{v}\right)^2}{2}}-\svalu \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{-A-u}{v}\right)^2}{2}} \right]+u^2+v^2\\
    &=\frac{v^2}{\eta}\left(\svall\pdfu-\svalu\pdfl\right)+u^2+v^2\\
    &=\varepsilon.
    \end{align}
\end{subequations}

因此，$ \widetilde{\alpha},\widetilde{\beta},\widetilde{\gamma} $是方程组\eqref{Eqn:P2P:SISO:Lower:MaxEntropy:Solution}的一组解。并且将其代入最大熵问题的拉格朗日函数的导数表达式，得
\begin{subequations}
    \begin{align}
    &\frac{\partial L}{\partial f^{\mathrm{ABG}}_X\left(x\right)}=\ln f^{\mathrm{ABG}}_X\left(x\right) + 1 + \widetilde{\alpha} +\widetilde{\beta} x + \widetilde{\gamma} x^2\\
    &=\ln e^{-1-\left(\frac{u^2}{2v^2}+\ln{ \eta v\sqrt{2\pi}}-1\right)-\left(-\frac{u}{v^2}\right)x-\frac{1}{2v^2}x^2} + 1 + \tgA \tgB x + \frac{x^2}{2v^2}\\
    &= -\tgG x^2 +\frac{u}{v^2}x-\frac{u^2}{2v^2}-\ln{\eta v\sqrt{2\pi}}+\ln{\eta v\sqrt{2\pi}}+\frac{u^2}{2v^2} \tgB x + \frac{x^2}{2v^2}\\
    &=0.
    \end{align}
\end{subequations}

所以$ f^{\mathrm{TG}}_X\left(x\right) $是满足约束的最大熵分布。